## anonymous 5 years ago solve the first order differential equation: y-x(dy/dx)=(dy/dx)y^2e^y

1. anonymous

Rearrange the equation to put it into a standard form,$y-(x^2+y^2e^y)\frac{dy}{dx}=0$which is in the form, $M(x,y)+N(x,y)\frac{dy}{dx}=0$We check to see if it's exact. We need to ensure,$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} \rightarrow 1=1$which is true. Now,

2. anonymous

the potential function,$\Psi(x,y)$has that$\frac{\partial \Psi}{\partial x}=M=y$and$\frac{\partial N}{\partial x}=N=-x^2-y^2e^y$

3. anonymous

Integrate the first partial:$\Psi(x,y)=xy+c(y)$and then take the partial derivative of this result with respect to y,$\frac{\partial \Psi}{\partial y}=x+c'(y)$so that we can identify it with $\frac{\partial \Psi}{\partial y}=-x^2-y^2e^y$

4. nowhereman

Above you were saying $\frac{∂N}{∂x} = 1$ while indeed it is $-2x$

5. anonymous

(note a mistake above dN/dx should be dPsi/dy)

6. nowhereman

I mean where you say "We check to see if it's exact."

7. anonymous

Oh - thanks nowhereman - typo with x^2...it should just be x

8. nowhereman

right :-)

9. anonymous

great - now the whole thing's polluted.

10. anonymous

$\frac{\partial \Psi}{\partial y}=-(x+ye^y)=x+c'(y)$

11. anonymous

so that$c'(y)=-2x-ye^y \rightarrow c(y)=-2xy-e^y(y^2-2y+2)$where the last integral is doable using a few integration by parts. Sub. this back in for our solution Psi to give,$\Psi(x,y)=xy-2xy-e^y(y^2-2y+2)$$=-xy-e^y(y^2-2y+2)+c=K$

12. anonymous

where c is the constant of integration, and K is a constant due to the fact that this function Psi satisfies$\frac{d \Psi(x,y)}{dx}=\frac{\partial \Psi}{\partial x}\frac{dx}{dx}+\frac{\partial \Psi}{\partial y}\frac{dy}{dx}=\Psi_{x}+\Psi_y \frac{dy}{dx}=0$which is the form of the equation we had to solve in the beginning.

13. anonymous

Are you seeing any mistakes, nowhereman?

14. anonymous

So $\Psi(x,y)=-xy-e^y(y^2-2y+2)=c$for c, some constant.